# Linear algebra

## Linear Function

For any 2 point A(0,0) , B($x_{o},y_{o}$ ) a straight line can be drawn . The slope of the line is

$m={\frac {y_{o}}{x_{o}}}$

Therefore,

$y_{o}=mx_{o}$

For a point C($x,y$ ) on the same line of the slope m

$m={\frac {y-y_{o}}{x-x_{o}}}$
$y=y_{o}+m(x-x_{o})$

## Linear Equations

Has general form

 $Ax=0$ $x=0$ $Ax=C$ $x={\frac {C}{A}}$ $Ax+B=C$ $x={\frac {C-B}{A}}$ $Ax+By=C$ $x={\frac {C}{A}}$ with $y=0$ $y={\frac {C}{B}}$ with $x=0$ ## System of Linear Equations

### Example

$2x+y=11\,$
$-4x+3y=13\,$

### Roots of system of equations

#### Variable Elimination

$2x+y=11\,$
$-4x+3y=13\,$

Multiply 1st row by 2 add to 2nd row

$4x+2y=22\,$
$-4x+3y=13\,$

Yields

$5y=35\,$  => $y=7$

Multiply 1st row by -3 add to 2nd row

$-6x-3y=-33\,$
$-4x+3y=13\,$

Yields

$-10x=-20\,$  => $x=2$

#### Substitution

If you get a system of equations that looks like this:

$2x+y=11\,$
$-4x+3y=13\,$

You can switch around some terms in the first to get this:

$y=-2x+11\,$

Then you can substitute that into the bottom one so that it looks like this:

$-4x+3(-2x+11)=13\,$
$-4x-6x+33=13\,$
$-10x+33=13\,$
$-10x=-20\,$
$x=2\,$

Then, you can substitute 2 into an x from either equation and solve for y. It's usually easier to substitute it in the one that had the single y. In this case, after substituting 2 for x, you would find that y = 7.

#### Determinant

If you get a system of equations that looks like this:

$2x+y=11\,$
$-4x+3y=13\,$

Solve for y

$x+{\frac {1}{2}}y={\frac {11}{2}}\,$
$x+{\frac {3}{-4}}y={\frac {13}{-4}}\,$
$y({\frac {1}{2}}-{\frac {3}{-4}})={\frac {11}{2}}-{\frac {13}{-4}}$
$y={\frac {{\frac {11}{2}}-{\frac {13}{-4}}}{{\frac {1}{2}}-{\frac {3}{-4}}}}$

Solve for x

$2x+y=11\,$
${\frac {-4}{3}}x+y={\frac {13}{3}}\,$
$x(2-{\frac {-4}{3}})=11-{\frac {13}{3}}$
$x={\frac {11-{\frac {13}{3}}}{2-{\frac {-4}{3}}}}$

#### Find Roots of System of Linear Equations systematically

$A_{1}x+B_{1}y=C_{1}$
$A_{2}x+B_{2}y=C_{2}$

Eliminate variable x

$x+{\frac {B_{1}}{A_{1}}}y={\frac {C_{1}}{A_{1}}}$
$x+{\frac {B_{2}}{A_{2}}}y={\frac {C_{2}}{A_{2}}}$

$y({\frac {B_{1}}{A_{1}}}-{\frac {B_{2}}{A_{2}}})={\frac {C_{1}}{A_{1}}}-{\frac {C_{2}}{A_{2}}}$

Solve for y

$y={\frac {{\frac {C_{1}}{A_{1}}}-{\frac {C_{2}}{A_{2}}}}{{\frac {B_{1}}{A_{1}}}-{\frac {B_{2}}{A_{2}}}}}$

Eliminate variable y

${\frac {A_{1}}{B_{1}}}x+y={\frac {C_{1}}{B_{1}}}$
${\frac {A_{2}}{B_{2}}}x+y={\frac {C_{2}}{B_{2}}}$

$x({\frac {A_{1}}{B_{1}}}-{\frac {A_{2}}{B_{2}}})={\frac {C_{1}}{B_{1}}}-{\frac {C_{2}}{B_{2}}}$
$x={\frac {{\frac {C_{1}}{B_{1}}}-{\frac {C_{2}}{B_{2}}}}{{\frac {A_{1}}{B_{1}}}-{\frac {A_{2}}{B_{2}}}}}$